Alexander Bass
This note is one of many taken during 2025

Partial sums of the Geometric Series+

So, the geometric series: if you’ve taken some Calculus, you know of it.

r=1+x+x2+x3+… r = 1 + x + x^2+x^3+ \dots

Solve for rr

r=1+rx r = 1 + rx r−rx=1 r-rx = 1 r(1−x)=1 r(1-x) = 1 r=11−x r = \frac{1}{1-x}

And there’s the formula the textbook gave ya'

∑j=0∞xj=11−x \sum_{j=0}^{\infty} x^j = \frac{1}{1-x}

But what about the partial sums?

∑j=0nxj=? \sum_{j=0}^{n} x^j = ?

Decompose into difference of sums

∑j=0nxj=∑j=0∞xj−∑j=n+1∞xj \sum_{j=0}^{n} x^j = \sum_{j=0}^{\infty} x^j - \sum_{j=n+1}^{\infty} x^j

We know the value of the left sum, and the right sum can be found similarly.

h=xn+1+xn+2+xn+3+… h = x^{n+1} + x^{n+2} + x^{n+3}+ \dots h=xn+1+x(xn+1+xn+2+… ) h = x^{n+1} + x(x^{n+1} + x^{n+2} + \dots) h=xn+1+x(h) h = x^{n+1} + x(h) h=xn+11−x h = \frac{x^{n+1}}{1-x}

So the partial sum can be found

∑j=0nxj=11−x−xn+11−x=1−xn+11−x \sum_{j=0}^{n} x^j = \frac{1}{1-x} - \frac{x^{n+1}}{1-x} = \frac{1-x^{n+1}}{1-x}

Some alternate forms are common

∑j=0n−1xj=1−xn1−x=xn−1x−1 \sum_{j=0}^{n-1} x^j = \frac{1-x^{n}}{1-x} = \frac{x^n-1}{x-1}

Also note that, now that the upper bound of the sum is not infinity, the sum is defined for values ∣x∣≥1 \lvert x \rvert \ge 1

Partial Extended-Geometric Sums

This also applies to series like

x1+x4+x7+x10+… x^1 + x^4 + x^7 + x^{10} + \dots

or in general terms where a>b≥0a > b \ge 0

r=xa+b+x2a+b+x3a+b+… r = x^{a+b} + x^{2a+b} + x^{3a+b} + \dots

Finding a closed form for this is not too much more difficult. First find the closed for to the simpler

h=xa+x2a+x3a+… h = x^{a} + x^{2a} + x^{3a} +\dots h=xa(1+xa+x2a+… ) h = x^{a}(1 + x^{a} + x^{2a} +\dots) h=xa+xah h = x^{a} + x^a h h=xa1−xa h =\frac{ x^{a}}{1-x^a}

Now the more complex form can be found

r=xa+b+x2a+b+x3a+b+… r = x^{a+b} + x^{2a+b} + x^{3a+b} + \dots r=xa+b+x2a+b+x3a+b+… r = x^{a+b} + x^{2a+b} + x^{3a+b} + \dots r=xb(xa+x2a+x3a+… ) r = x^b (x^{a} + x^{2a} + x^{3a} + \dots) r=xbh=xa+b1−xa r= x^bh = \frac{x^{a+b}}{1-x^a} ∑j=1∞xaj+b=xa+b1−xa \sum_{j=1}^{\infty} x^{aj+b} = \frac{x^{a+b}}{1-x^a}

Following similar reasoning as the regular geometric series, the partial form is found.

m=xan+b+xa(n+1)+b+xa(n+2)+b+… m = x^{an+b} + x^{a(n+1)+b} + x^{a(n+2)+b} + \dots m=xb(xan+xa(n+1)+xa(n+2)+… ) m = x^b( x^{an} + x^{a(n+1)} + x^{a(n+2)} + \dots) m=xb(xan+xa(n+1)+xa(n+2)+… ) m = x^b ( x^{an} + x^{a(n+1)} + x^{a(n+2)} + \dots)

Pause solving mm, introduce new equation, tt

t=xan+xa(n+1)+xa(n+2)+… t = x^{an} + x^{a(n+1)} + x^{a(n+2)} + \dots t=xan+xan+1a+xan+2a+… t = x^{an} + x^{an+1a} + x^{an+2a} + \dots t=xan(1+x1a+x2a+… ) t = x^{an}(1 + x^{1a} + x^{2a} + \dots)

Pause solving tt, introduce new equation oo

o=xa+x2a+x3a+… o = x^{a} + x^{2a} + x^{3a} + \dots o=xa(1+o) o = x^{a}(1 + o) o=xa1−xa o = \frac{x^a}{1-x^a}

Resume solving tt

t=xan(1+x1a+x2a+… ) t = x^{an}(1 + x^{1a} + x^{2a} + \dots) t=xan(1+xa1−xa) t = x^{an}\left(1 + \frac{x^a}{1-x^a}\right) t=xan11−xa t = x^{an}\frac{1}{1-x^a}

Resume solving mm

m=xb(xan+xa(n+1)+xa(n+2)+… ) m = x^b ( x^{an} + x^{a(n+1)} + x^{a(n+2)} + \dots) m=xan+b1−xa m = \frac{x^{an+b}}{1-x^a}

Thus

∑j=n∞xaj+b=xan+b1−xa \sum_{j=n}^{\infty} x^{aj+b} = \frac{x^{an+b}}{1-x^a}

So

∑j=1n−1xaj+b=∑j=1∞xaj+b−∑j=n∞xaj+b \sum_{j=1}^{n-1} x^{aj+b} = \sum_{j=1}^{\infty} x^{aj+b} - \sum_{j=n}^{\infty} x^{aj+b} ∑j=1n−1xaj+b=xa+b1−xa−xan+b1−xa=xa+b−xan+b1−xa=xa+b(1−xa(n−1))1−xa \sum_{j=1}^{n-1} x^{aj+b} = \frac{x^{a+b}}{1-x^a} - \frac{x^{an+b}}{1-x^a} = \frac{x^{a+b} - x^{an+b}}{1-x^a} = \frac{x^{a+b}\left(1-x^{a(n-1)} \right)}{1-x^a}

For convenience, increment the value of nn by one.

∑j=1nxaj+b=xa+b(1−xan)1−xa \sum_{j=1}^{n} x^{aj+b} = \frac{x^{a+b}\left(1-x^{an} \right)}{1-x^a}

When 0≤b<a0 \le b < a

I’m not sure if any of that made any sense, but from my bit of verification, I believe it’s true.

As far as verification goes, it wouldn’t hurt to do so through Induction.

First the base case of n=1n=1

∑j=11xaj+b=xa+b \sum_{j=1}^{1} x^{aj+b} = x^{a+b} xa+b(1−xa)1−xa \frac{x^{a+b}\cancel{\left(1 - x^{a}\right)}}{\cancel{1-x^a}}

The base case is satisfied

The equation is valid if the following holds

xa+b(1−xan)1−xa+xa(n+1)+b=xa+b(1−xa(n+1))1−xa \frac{x^{a+b}\left(1-x^{an} \right)}{1-x^a} + x^{a(n+1)+b} = \frac{x^{a+b}\left(1-x^{a(n+1)} \right)}{1-x^a} xa+b(1−xan)+xa(n+1)+b(1−xa)=xa+b(1−xa(n+1)) x^{a+b}\left(1-x^{an} \right) + x^{a(n+1)+b}({1-x^a}) = x^{a+b}\left(1-x^{a(n+1)} \right) xa+b−xa(n+1)+b+xa(n+1)+b(1−xa)=xa+b−xa(n+2)+b x^{a+b} -x^{a(n+1)+b} + x^{a(n+1)+b}({1-x^a}) = x^{a+b}-x^{a(n+2) + b} xa+b−xa(n+1)+b+xa(n+1)+b−xa(n+2)+b=xa+b−xa(n+2)+b x^{a+b} -x^{a(n+1)+b} + x^{a(n+1)+b} - x^{a(n+2)+b} = x^{a+b}-x^{a(n+2) + b} xa+b−xa(n+1)+b+xa(n+1)+b−xa(n+2)+b=xa+b−xa(n+2)+b x^{a+b} \cancel{-x^{a(n+1)+b} + x^{a(n+1)+b}} - x^{a(n+2)+b} = x^{a+b}-x^{a(n+2) + b}

And so it has been proven.

This note is one of many taken during 2025